Answer to Question #116108 in Optics for AA

Question

Initially unpolarized light is incident on a series a polarizing sheets. You decide to investigate what percentage of the initial intensity is transmitted for a variety of configurations of the system. Give that percentage for each of the following situations.

a. Two polarizers with the second sheet's polarizing angle at 90 degrees with the first sheet's.
b. You add an additional polarizer in the middle of the configuration in part a, whose angle is halfway between the other two's angle
c. 4 polarizing sheets, each at an angle of 30 degrees with the previous.
d. We could obviously continue adding as many sheets as we like in this fashion. if we now have light polarized in the x direction, and we want to change its polarization by 90 degrees, what's the minimum number of polarizing sheets do we need in order to keep 90% of its intensity?

Answer 1

We know that when the polarizers are rotated for angle θ, the transmitted intensity will be

$$I_t=I_0\text{ cos}^2\theta.$$

This is all we need to solve the problem.

a. At 90° angle the percentage of the initial intensity transmitted is:

$$\frac{I_t}{I_0}=\text{ cos}^290^\circ\cdot100\%=0.$$

b. After the first polarizer we only have a half of the initial intensity:

$$I_1=0.5I_0.$$

After the first shift (after two sheets 45° to each other, before the third polarizer) we have

$$I_2=I_1\text{ cos}^245^\circ=0.5I_1=0.25I_0.\\$$

After the second shift (after the third polarizer):

$$I_3=I_2\text{ cos}^245^\circ=0.5I_2=0.125I_0,$$

or 12.5%.

c. According to the previous case, we have half initial intensity after the first sheet,

$$I_2=I_1\text{ cos}^230^\circ=0.75I_1=0.375I_0$$

after the second sheet,

$$I_3=I_2\text{ cos}^230^\circ=0.75I_2=0.28I_0$$

after the third one,

$$I_4=I_3\text{ cos}^230^\circ=0.75I_3=0.21I_0$$

after the four sheets shifted 30° to each other, we have 21% of the initial intensity.

d. From problems (b) and (c) we could see the pattern. After 3 and 4 sheets the final intensity respectively was

$$I_3=0.5I_0[\text{cos}^2\theta]^{3-1},\\I_4=0.5I_0[\text{cos}^2\theta]^{4-1},$$

therefore, the rule is

$$I_n=\frac{I_0[\text{cos}^2\theta]^{n-1}}{2}.$$

Since we need 90° shift, angle θ must comply with the rule

$$90^\circ=(n-1)\theta,$$

therefore, the transmitted intensity is

$$I_n=I_0\frac{\Big[\text{cos}^2\big(\frac{90^\circ}{n-1}\big)\Big]^{n-1}}{2}.$$

To keep 90% intensity, we must find n from the following equation:

$$\frac{I_n}{I_0}=0.9=\frac{\Big[\text{cos}^2\big(\frac{90^\circ}{n-1}\big)\Big]^{n-1}}{2}.$$

This equation does not have solution.